Problem: Subtract. $\dfrac{9}{8} - \dfrac{4}{10} = $
Solution: Before we can subtract our fractions, they need to have the same denominator. $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\frac{1}{10}$ $\dfrac{9}{8}$ $\dfrac{4}{10}$ $\dfrac{9}{8}-\dfrac{4}{10}$ Let's look at the multiples of each denominator and see which multiples they have in common. Denominator Multiples ${8}$ $8, 16, 24, 32, \underline{40}$ $10}$ $10, 20, 30, \underline{40}$ The least common denominator is ${40}$. Let's use multiplication to make each fraction have a denominator of $40$. ${\dfrac{9}{8}}=\dfrac{{9} \times {5}}{{8} \times {5}} = {\dfrac{45}{40}}$ $\dfrac{4}{10}}=\dfrac{4} \times 4}{10} \times 4} = {\dfrac16}40}}$ Now, we can subtract ${\dfrac{45}{40}} - \dfrac{16}{40}}$. $\dfrac{45}{40}$ $\dfrac{16}{40}$ $\dfrac{45}{40} - \dfrac{16}{40}$ $=\dfrac{{45}-16}}{40}$ $= \dfrac{29}{40}$ ${\dfrac{9}{8}} - \dfrac{4}{10}} = \dfrac{29}{40}$